Chapter 4.8, Problem 15P

To determine

To find how high will it go? When will it return to the ground?

Answer to Problem 15P

The projectile will go high at 490 meters and it will return to the ground in 20 seconds.

Explanation of Solution

Given information: Initial speed of projectile is $98m/s$ .

Calculation: The required solution is obtained as,

Use the projectile equation, $h=vt-4.9t^2$ .

When $h=0$ the projectile return to the ground. Solve for t.

  $\begin{array}{l}h=vt-4.9t^2\\ 0=98t-4.9t^2\\ 4.9t^2-98t=0\\ t^2-20t=0\\ t\left(t-20\right)=0\\ t=0,20\end{array}$

Take $t=20$ sec.

Find the maximum height by substituting $t=10$ seconds into the equations.

  $\begin{array}{l}h=vt-4.9t^2\\ h=98\left(10\right)-4.9{\left(10\right)}^2\\ h=980-490\\ h=490\end{array}$

Hence, the projectile will go high at 490 meters and it will return to the ground in 20 seconds.